@DavidESpeyer The inequality is false for general symmetric real $A,B$. A simple example is as follows. Let $P,Q$ be two orthogonal projections in $\mathbb R^2$ so that $0<\|PQ\|_F<\frac 12$ and let $P',Q'$ be the complementary orthogonal projections. Then $U=(P-P')(Q-Q')$ is a product of two reflections, i.e., a rotation and we can easily arrange it to be a rotation by an irrational multiple of $\pi$. Then for every unit vector $x$, the orbit $U^mx:m\ge 0$ is dense on the unit circle.

In particular, if $x$ is a unit vector such that $Qx=x$ and $y$ is a unit vector such that $Py=y$, then we can find $m$ so that $U^mx\approx y$ and, thereby, $\|PU^mQ\|\approx 1$ even for the operator norm.

Now choose $a\in(0,1)$ so close to $1$ that the same holds for $U_a=(P-aP')(Q-aQ')$ instead of $U$. At last, put $A=P-aP', B=Q-aQ'$ and choose so large power $N$ that $A^n\approx P$ and $B^n\approx Q$ for all $n\ge N$. Then
$$
\|A^N(AB)^mB^N\|_F\approx \|PU_a^m Q\|_F\ge \|PU_a^mQ\|\approx 1
\\
>\frac 12>\|PQ\|_F\approx \|A^{N+m}B^{N+m}\|_F
$$
I cannot pull this trick with positive definite $A,B$ though, so that case still remains to be investigated.

**Edit** Here is a counterexample in $\mathbb R^3$ for the positive semi-definite case. Let $x,y$ be 2 orthogonal unit vectors in $\mathbb R^3$ and let $z$ be the unit vector that is linearly independent with $x,y$ and makes an angle $\pi/3$ with both of them (the exact angle value is not important as long as it is strictly smaller than $\pi/2$). For unit vectors $u,v$, let $P_u$ be the orthogonal projection to the line spanned by $u$ and let $P_{uv}$ be the orthogonal projection to the plane spanned by $u$ and $v$.

Note that $(P_{xz}P_{yz})^m\to P_z$ as $m\to\infty$. Choose a large $m$ so that $(P_{xz}P_{yz})^m\approx P_z$ with high precision. Now let $x'$ be the unit vector orthogonal to $x$ in the plane spanned by $x,z$ and let $y'$ be the unit vector orthogonal to $y$ in the plane spanned by $y,z$. We have $P_{xz}=P_x+P_{x'}$ and similarly for $P_{yz}$. Choose $a\in (0,1)$ so close to $1$ that the operators $A=P_x+aP_{x'}$ and $B=P_y+aP_{y'}$ still satisfy $(AB)^m\approx P_z$. Now choose $N$ so that for all $n\ge N$, we have $A^n\approx P_x$, $B^n\approx P_y$. Then
$$
\|A^N(AB)^mB^N\|_F\approx\|P_xP_zP_y\|_F=\frac 14
\\
>0=\|P_xP_y\|_F\approx \|A^{N+m}B^{N+m}\|_F\,.
$$

Thus there is no hope for the general version of the problem even for positive definite matrices. However, for *positive definite matrices* there are two statements that are true and not too hard to prove:

(1) The conjecture holds in $\mathbb R^2$ (so my 3-dimensional counter-example is a minimal one)

(2) If $\alpha_j\ge 0, \sum_j \alpha_j=1$, then for any positive definite $A,B$ in any dimension, we have
$$
\|C_1\dots C_m\|_F\le \|AB\|_F
$$
where each $C_j$ is either $A^{\alpha_j}B^{\alpha_j}$ or $B^{\alpha_j}A^{\alpha_j}$. This takes care of the products like $AB^2AB$, but not of $A^2BAB$, say. I wonder if this restricted type of products is actually the best we can confirm the conjecture for.